You are given 1.226 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl and O2,

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),
and 246 mL of O2 is collected over water at 24 °C. The total pressure of the gases in the collection flask is 755 torr. What is the weight percentage of KClO3 in the sample?
The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 24 °C is 22.4 torr.

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

Use PV = nRT and solve for mols O2 at the conditions listed.
Using the coefficients in the balanced equation, convert mols O2 to mols KClO3.
Convert mols KClO3 to grams by g = mols x molar mass
Then %KClO3 = (mass KClO3/mass sample)*100 = ?

To calculate the weight percentage of KClO3 in the sample, we need to determine the moles of O2 collected, which will give us the moles of KClO3 decomposed. Then we can calculate the weight of KClO3 in the sample.

Step 1: Calculate the moles of O2 collected
PV = nRT

Where:
P = Total pressure of gases (in torr) - vapor pressure of water
V = Volume of O2 collected (in liters) - 246 mL = 0.246 L
n = Moles of gas
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin - 24 °C = 24 + 273 = 297 K

Convert the total pressure to atm by subtracting the vapor pressure of water:
P = 755 torr - 22.4 torr = 732.6 torr = 732.6/760 atm

PV = nRT
n = PV/RT
n = (732.6/760) * 0.246 / (0.0821 * 297)

n = 0.0088 moles of O2

Step 2: Calculate the moles of KClO3 decomposed
According to the balanced equation:
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

2 moles of KClO3 decompose to produce 3 moles of O2.
Therefore, the moles of KClO3 decomposed will be: (2/3) * 0.0088 = 0.00587 moles

Step 3: Calculate the weight of KClO3 in the sample
The molar mass of KClO3 is given as 122.55 g/mol.
Therefore, the weight of KClO3 in the sample is: 0.00587 moles * 122.55 g/mol = 0.718 g

Step 4: Calculate the weight percentage of KClO3 in the sample
Weight percentage = (weight of KClO3 / weight of mixture) * 100
Weight percentage = (0.718 g / 1.226 g) * 100

Weight percentage of KClO3 in the sample is approximately 58.62%.

To find the weight percentage of KClO3 in the sample, we need to determine the amount of KClO3 and KCl present in the mixture.

1. Calculate the number of moles of O2 collected:
We know that the O2 gas collected over water at 24 °C has a pressure of 755 torr. However, we need to correct for the vapor pressure of water at that temperature (22.4 torr). Therefore, subtract the vapor pressure of water from the total pressure: 755 torr - 22.4 torr = 732.6 torr.

Convert the pressure to atm: 732.6 torr / 760 torr/atm = 0.963 atm.

To find the number of moles of O2, we can use the ideal gas law. Rearranging the equation, we have:
PV = nRT,
where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Since we collected the gas over water, the volume of the collected gas is the same as the volume of the gas in the flask: 246 mL = 0.246 L.
Thus, the number of moles of O2 is: n(O2) = (0.963 atm)(0.246 L) / (0.0821 L·atm/mol·K)(24 + 273 K) = 0.0107 mol.

2. Calculate the number of moles of KClO3:
From the balanced chemical equation, we know that for every 3 moles of O2 released, there are 2 moles of KClO3. Therefore, the number of moles of KClO3 can be calculated using the ratio:
n(KClO3) = (2/3) * n(O2) = (2/3) * 0.0107 mol = 0.00713 mol.

3. Calculate the mass of KClO3:
Using the molar mass of KClO3 (122.55 g/mol), we can calculate the mass of KClO3 in the sample:
Mass(KClO3) = n(KClO3) * Molar Mass(KClO3) = 0.00713 mol * 122.55 g/mol = 0.874 g.

4. Calculate the weight percentage of KClO3 in the sample:
Weight percentage of KClO3 = (Mass(KClO3) / Mass of the sample) * 100%.

The mass of the sample is given as 1.226 g. Therefore:
Weight percentage of KClO3 = (0.874 g / 1.226 g) * 100% = 71.4%.

So, the weight percentage of KClO3 in the sample is 71.4%.